Question

Can someone help me please.

Let f(x) 5x^2-21x-20/3x^2-13x+12


This function has: 1) A y intercept at the point


2) x intercepts at the point(s)


3) Vertical asymptotes of
and
(Give the left vertical asymptote first)

4) Horizontal asymptote of

Answer 1

1) To find the y-intercept, we plug in x=0 into the function:

f(0) = (5(0)^2 - 21(0) - 20) / (3(0)^2 - 13(0) + 12) = -20/12 = -5/3

So the y-intercept is at the point (0, -5/3).

2) To find the x-intercepts, we set f(x) = 0 and solve for x:

5x^2 - 21x - 20 = 0
(5x + 4)(x - 5) = 0

So the x-intercepts are at the points (-4/5, 0) and (5, 0).

3) To find the vertical asymptotes, we set the denominator equal to zero and solve for x:

3x^2 - 13x + 12 = 0
(3x - 4)(x - 3) = 0

So the vertical asymptotes are at x = 4/3 and x = 3.

4) To find the horizontal asymptote, we look at the degrees of the numerator and denominator. Since the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is y = 0.

Therefore, the answers are:
1) y-intercept at (0, -5/3)
2) x-intercepts at (-4/5, 0) and (5, 0)
3) vertical asymptotes at x = 4/3 and x = 3
4) horizontal asymptote at y = 0.